What is the common formula of “Prussian blue is” and “Turn bull's blue”?

 Prussian blue is a dark blue pigment, the another name of Prussian blue is Berlin blue or Paris blue. It is formed by oxidation of ferrous ferrocyanide salts with FeCl₃.

Colour of Prussian blue and turnbull’s blue is originated by charge transfer when electronic transition occurs from one part of the Complex to other part i.e such type are also called internal redox reaction (Charge transfer from metal  to metal). Intensity of clour in such type transition is very high as they do not require follow any selection rule.

      4FeCl₃​+3K₄​[Fe(CN)₆​]→Fe₄​[Fe(CN)₆​]₃​+12KCl
                                  OR  “Prussian blue is” or “Turn bull's blue”
       FeCl₃​+K₄​[Fe(CN)₆​]→KFe[Fe(CN)₆​]+3KCl

                                  OR
       FeSO₄​+K₃​[Fe(CN)₆​]→KFe[Fe(CN)₆​]+K₂​SO₄​
                                            “Turn bull's blue”

Related Question:

Although both [Mn(H2O)6]2+ and [FeF6]3- have a d5 configuration and high-spin complexes. But the dilute solutions of Mn2+ and Fe +3 complexes are therefore colorless. Why?

Which of the Complex of the following pairs has the highest value of CFSE?
Colour of Complexes due to charge transfer:
Why violet colour of [Ti(H2O)6]Cl3 disapear (colourless) on heating heating ? 
Why [Ni(CN)4]-2 is colourless while [Ni(H2O)4]-2 although both have +2 oxidation state and 3d*8 configuration ?
Why [FeF6]3– is colourless whereas [CoF6]3– is coloured ? 
Why Fe(CO)5 is colourless while Fe(bipy)(CO)3 is intensely purple in colour ?
Why all the tetrahedral Complexes are high spin Complexes ?

 

Why Cis-But-2-ene posseses dipole moment while Trans-But-2-ene have zero dipole moment ?

  We know that But -2- ene It exists in two forms Cis-But-2-ene and Trans-But-2-ene

                      

SOLUTION: The Trans isomer is symmetrical with the 2-methyl groups in anti position. So the bond dipoles the two Me– C bonds acting in opposition cancel each other and result in a zero dipole. Where as in cis isomer the dipoles do not cancel each other resulting in a net dipole

 

Related Question

 

What is "Pouling Equation" and How to calculate % ionic character of covalent bond ?:

What is" Haney and Smith Scale" and How to use calculation of % ionic character of covalent bond ?

How to calculate the & ionic character in H-X bond of Halogen acids.Given that electronegativity of H ,F, Cl, Br. and I are 2.1, 4.0, 3.0, 2.8,and 2.5 respectively.

What is "Pouling Equation" and How to calculate % ionic character of covalent bond ?

Calculate the % of ionic character of a bond having length is 0.83 Å and 1.82 D as its observed dipole moment ?

Calculate the % of ionic character of a bond having length 0.92 Å and 1.91 D as its observed dipole moment ?.

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Calculate the % of ionic character of a bond having length 0.92 Å and 1.91 D as its observed dipole moment ?.

   To calculate μ considering 100% ionic bond:

                      = 4.8 × 10^–10 × 0.92 × 10^–8esu cm

                      = 4.8 × 0.83 × 10^–18 esu cm = 4.416 D

                         ∴ % ionic character = 1.91/4.416× 100 = 43.25%

 

Related Questions:

What is" Haney and Smith Scale" and How to use calculation of % ionic character of covalent bond ?

How to calculate the & ionic character in H-X bond of Halogen acids.Given that electronegativity of H ,F, Cl, Br. and I are 2.1, 4.0, 3.0, 2.8,and 2.5 respectively.

What is "Pouling Equation" and How to calculate % ionic character of covalent bond ?

Calculate the % of ionic character of a bond having length is 0.83 Å and 1.82 D as its observed dipole moment?

 

         

What is "Pouling Equation" and How to calculate % ionic character of covalent bond ?

The "Pouling Equation" of % Ionic Charater of covalent compound given as"

Related Questions:

What is" Haney and Smith Scale" and How to use calculation of % ionic character of covalent bond ?

How to calculate the & ionic character in H-X bond of Halogen acids.Given that electronegativity of H ,F, Cl, Br. and I are 2.1, 4.0, 3.0, 2.8,and 2.5 respectively.

Calculate the % of ionic character of a bond having length is 0.83 Å and 1.82 D as its observed dipole moment?

 

 

 

Calculate the % of ionic character of a bond having length is 0.83 Å and 1.82 D as its observed dipole moment?

Pauling used dipole moment for depicting percentage of polarity and ionic character of the bond. According to Pauling, a bond can never be 100% ionic.

To calculate μ considering 100% ionic bond

                      = 4.8 × 10^–10 × 0.83 × 10^–8esu cm

                      = 4.8 × 0.83 × 10^–18 esu cm = 3.984 D

                         ∴ % ionic character = 1.82/3.984 × 100 = 45.68%

(#) When electronegativity difference between two atoms is 2.1, there is 50% ionic character in the bond.

(#) When electronegativity difference is zero (identical atoms), the bond will be 100% covalent.

 

Related Questions:

What is" Haney and Smith Scale" and How to use calculation of % ionic character of covalent bond ?

How to calculate the & ionic character in H-X bond of Halogen acids.Given that electronegativity of H ,F, Cl, Br. and I are 2.1, 4.0, 3.0, 2.8,and 2.5 respectively.

What is "Pouling Equation" and How to calculate % ionic character of covalent bond ?